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Math Help - More rational/irrational junk

  1. #1
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    More rational/irrational junk

    Here we go again:

    If a,b are real numbers with a < b prove:

    1) There is a rational r within (a,b).
    I am given the hint to start by selecting an interger q > 0 such that
    1/q < b-a. I am not sure how it applies.
    I am trying to go the route a < q < b and somehow show that q is a rational.

    2) There is a irrational c within (a,b).
    Says to look at part 1)

    3) (a,b) contains infinitely many rationals and irrationals.

    Seems like once it is shown to me how to do it, I can grasp the concept, but it seems kind of difficult to assign homework that is not lectured on how to do. Oh well, thanks in advance MHF people. Off to class.
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  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by Sinuyen View Post
    Here we go again:

    If a,b are real numbers with a < b prove:

    1) There is a rational r within (a,b).
    I am given the hint to start by selecting an interger q > 0 such that
    1/q < b-a. I am not sure how it applies.
    I am trying to go the route a < q < b and somehow show that q is a rational.

    2) There is a irrational c within (a,b).
    Says to look at part 1)

    3) (a,b) contains infinitely many rationals and irrationals.

    Seems like once it is shown to me how to do it, I can grasp the concept, but it seems kind of difficult to assign homework that is not lectured on how to do. Oh well, thanks in advance MHF people. Off to class.
    1) WLOG, let 0<a<b. From the hint, it follows that 1 < qb-qa. then qa+1 < qb

    qa \in \mathbb{R}^+. thus, there is an integer m such that m-1 \leq qa < m

    in particular, m\leq qa+1 < qb. from here, qa < m < qb implying that a < \frac{m}{q} < b


    for other items, use density property..
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